Please check my math!

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davedottcom

davedottcom

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(5) Single phase loads of 22 amps each on a 120/240 3-phase panel (w/ 208 hi-leg)

The 30 amp 2-pole breakers will be placed as follows:

#1 A-B
#2 C-A
#3 B-C
#4 A-B
#5 C-A

22 x 240 = 26,400 watts

3-phase: (240 x 1.73) = 415
26,400 / 415 = 63.6 Amps
Continuos load so: 63.6 x 125% = 79.5

(3) #4 copper feeders to the panel?

It's been awhile since I tried this stuff! I hope I'm at least close! :-?
 
You have a slight problem with your calculation in that the total load is not balanced and the method of calculation assumes a balanced 3? load. The easiest approach to this problem is to add a balancing phantom 6th load (think of it as a future addition)...

6 ? 22A ? √3 = 76.2A

This value represents the amount of current on the one leg connected to four loads. The other legs are connected to only three each and will carry less current.
 
Hmmm... 98 people viewed this thread & only 1 reply!
I guess I'm not the only one confused! :grin:

Thanks Smart $
 
davedottcom said:
Hmmm... 98 people viewed this thread & only 1 reply!
I guess I'm not the only one confused! :grin:

Thanks Smart $
Click to expand...

If it is a high number of replies you want ask a stupid question ( like ground up or down ) then they will beat down the door to get in a reply.:D
I like Smarts way of going at the problem by adding a phantom load in the calc. It made more sense then anyone else's idea.:D
 
acrwc10 said:
I like Smarts way of going at the problem by adding a phantom load in the calc. It made more sense then anyone else's idea.:D
Click to expand...

Smart $'s idea was definitely good although I don't understand what it means to add the 6th phantom load into the equation. I think I like the other people's replies better since I can grasp them easier. :)
 
steelersman said:
Smart $'s idea was definitely good although I don't understand what it means to add the 6th phantom load into the equation. I think I like the other people's replies better since I can grasp them easier. :)
Click to expand...

Glad to see some one gets my humor.:D
 
Smart $ said:
This value represents the amount of current on the one leg connected to four loads. The other legs are connected to only three each and will carry less current.
Click to expand...
With line-to-line loads, wouldn't that be two legs with the greater load and one with the lesser?
 
LarryFine said:
With line-to-line loads, wouldn't that be two legs with the greater load and one with the lesser?
Click to expand...
For an individual line to line load, yes... but feeder current here is the accumulation of several line to line loads. Count "letters" in the OP...

4 A's
3 B's
3 C's
 
Smart $ said:
For an individual line to line load, yes... but feeder current here is the accumulation of several line to line loads. Count "letters" in the OP...

4 A's
3 B's
3 C's
Click to expand...
I drew it out. Right you are!
 
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